The third line of the Balmer series spectrum of a hydrogen-like ion of atomic number `Z` equals to `108.5 nm`. The binding energy of the electron in the ground state of these ions is `E_(n)`. Then

Options

A. `Z = 2`
B. `E_(B) = 54.4 eV`
C. `Z = 3`
D. `E_(B) = 122.4 eV`

Answer

A, B

Explanation

For the third line of Balmer series, `n_(1) = 2, n_(2) = 5`
`:. (1)/(lambda) = RZ^(2) ((1)/(n_(1)^(2)) - (1)/(n_(2)^(2))) = RZ^(2) ((1)/(2^(2)) - (1)/(5^(2))) = (21 RZ^(2))/(100)`
`E = - 13.6 eV`
`Z^(2) xx (21)/(100) = (hc)/(lambda) = (1242 eV nm)/(108.5 nm)`
`Z^(2) = (1242 xx 100)/(108.5 xx 21 xx 13.6) = 4 implies Z = 2`
Binding energy of an electron in the ground state of hydrogen - like ion `= 13.6 Z^(2)//n^(2) = 54.4 eV (n = 1)`.